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For the purposes of the question, define $\exp^n x$ as the $n$ times iteration of $\exp$ (e.g $\exp^2 x = \exp \exp x$) for $n\geq1$ and $\exp^0 x = x$.

Let $f,g:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that $f,g$ are strictly increasing and unbounded above. If for all $n \geq 0$,

$$\exp^n f(x) \sim \exp^n g(x) ,\ \ x \to +\infty $$

does it follow that $f(x) = g(x)$ for sufficiently large $x$? If so, can the hypotheses be weaked? If not, are there hypotheses which can be added (e.g continuity)?

The motivation is this: even if $f \sim g$ as $x \to +\infty$, $f$ and $g$ are not really equivalent in the sense that for any well behaved function $h$, we don't necessarily have $h \circ f \sim h \circ g$ at $+\infty$.

For example, if $f \sim g$ but $|f - g|$ does not go to zero as $x \to +\infty$, $\exp f$ is not $\sim \exp g$.

Therefore, "highly" equivalent functions satisfying the hypotheses must have differences which tend to $0$. That's not sufficient, though. For example, $f(x) = x$, $g(x) = x + \frac 1x$ fails for $n=2$.

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MathematicsStudent1122
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1 Answers1

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This is not true; you can construct counterexamples by diagonalizing. Here's one specific way to do it. First, note that for any $n$, there exists $\delta_n\in(0,1)$ such that if $x\in[n,n+1]$ then $\exp^m(x+\delta_n)-\exp^m(x)<1$ for all $m\leq n$ (this is follows from uniform continuity of the functions $\exp^m$ on the compact set $[n,n+2]$). Now take $f(x)=x$ and define $g(n)=n+\delta_n$ for each $n\in\mathbb{N}$ and interpolate $g$ linearly for non-integer arguments. We then have $\exp^n(g(x))-\exp^n(f(x))<1$ whenever $x\geq n$. It follows that $\exp^n(g(x))\sim\exp^n(f(x))$ for all $n$. But $g(x)>f(x)$ for all $x$, since each $\delta_n$ is positive.

More generally, given any $f(x)$, you can similarly construct a counterexample $g(x)$ by arranging so that on $[n,n+1]$, $\exp^m(g(x))\in(\exp^m(f(x)),\exp^m(f(x))+1]$ for all $m\leq n$.

Eric Wofsey
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